Question

The phase difference between two waves $x_1 = A \; \sin (\omega t + \frac{\pi}{6} )$ and $x_2 = A \cos \omega t $ is

• A. $\frac{\pi}{2}$
• B. $\frac{\pi}{3}$
• C. $\frac{\pi}{6}$
• D. $\pi$

Answer 1

Answer: (B)

Here, the equation of two waves
$ x_{1} =A\, sin \left(\omega t+\frac{\pi}{6}\right)$ and
$x_{2} = A \,cos \left(\omega t\right) $
$\because sin \left(\theta +\frac{\pi}{2}\right) = cos\, \theta $
Hence, $x_{2} = A \,sin \left(\omega t +\frac{\pi}{2}\right)$
Now, the phase difference
$\phi = \left(\omega t +\frac{\pi}{2}\right) - \left(\omega t +\frac{\pi}{6}\right) $
$\Rightarrow \phi = \frac{\pi}{2} -\frac{\pi}{6} = \frac{\pi}{3} rad $
Hence, the phase difference between the waves is $\frac{\pi}{3}$ rad.